Clustering and classification
Load the data
Load the data from the MASS package and explore the structure and dimensions
library(dplyr)##
## Attaching package: 'dplyr'## The following objects are masked from 'package:stats':
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## filter, lag## The following objects are masked from 'package:base':
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## intersect, setdiff, setequal, unionlibrary(ggplot2)
library(MASS)##
## Attaching package: 'MASS'## The following object is masked from 'package:dplyr':
##
## selectdata("Boston")
str(Boston)## 'data.frame': 506 obs. of 14 variables:
## $ crim : num 0.00632 0.02731 0.02729 0.03237 0.06905 ...
## $ zn : num 18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
## $ indus : num 2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
## $ chas : int 0 0 0 0 0 0 0 0 0 0 ...
## $ nox : num 0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
## $ rm : num 6.58 6.42 7.18 7 7.15 ...
## $ age : num 65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
## $ dis : num 4.09 4.97 4.97 6.06 6.06 ...
## $ rad : int 1 2 2 3 3 3 5 5 5 5 ...
## $ tax : num 296 242 242 222 222 222 311 311 311 311 ...
## $ ptratio: num 15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
## $ black : num 397 397 393 395 397 ...
## $ lstat : num 4.98 9.14 4.03 2.94 5.33 ...
## $ medv : num 24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...dim(Boston)## [1] 506 14The Boston dataset is a data frame with 506 rows and 14 columns. The columns present different housing values in the suburbs of Boston.
Graphical overview of data
pairs(Boston)
summary(Boston)## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08204 1st Qu.: 0.00 1st Qu.: 5.19 1st Qu.:0.00000
## Median : 0.25651 Median : 0.00 Median : 9.69 Median :0.00000
## Mean : 3.61352 Mean : 11.36 Mean :11.14 Mean :0.06917
## 3rd Qu.: 3.67708 3rd Qu.: 12.50 3rd Qu.:18.10 3rd Qu.:0.00000
## Max. :88.97620 Max. :100.00 Max. :27.74 Max. :1.00000
## nox rm age dis
## Min. :0.3850 Min. :3.561 Min. : 2.90 Min. : 1.130
## 1st Qu.:0.4490 1st Qu.:5.886 1st Qu.: 45.02 1st Qu.: 2.100
## Median :0.5380 Median :6.208 Median : 77.50 Median : 3.207
## Mean :0.5547 Mean :6.285 Mean : 68.57 Mean : 3.795
## 3rd Qu.:0.6240 3rd Qu.:6.623 3rd Qu.: 94.08 3rd Qu.: 5.188
## Max. :0.8710 Max. :8.780 Max. :100.00 Max. :12.127
## rad tax ptratio black
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38
## Median : 5.000 Median :330.0 Median :19.05 Median :391.44
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90
## lstat medv
## Min. : 1.73 Min. : 5.00
## 1st Qu.: 6.95 1st Qu.:17.02
## Median :11.36 Median :21.20
## Mean :12.65 Mean :22.53
## 3rd Qu.:16.95 3rd Qu.:25.00
## Max. :37.97 Max. :50.00There’s both continuous and binary variables in the data. Some of the varibles are highly correlated, either negatively, such as lower status of the population (lstat) and median value of owner-occupied homes in $1000 (medv), or positively, such as full-value property-tax rate per $10,000 (tax) and proportion of non-retail business acres per town (indus).
However, they are not all normally distributed or have the same variance which are the assumptions for LDA. So we need to scale the variables.
Standardization and categorize the crime rate
We scale the values to have a mean of 0 and standard deviation of 1.
boston_scaled <- scale(Boston)
boston_scaled <- as.data.frame(boston_scaled)
apply(boston_scaled,2, sd, na.rm=TRUE)## crim zn indus chas nox rm age dis rad
## 1 1 1 1 1 1 1 1 1
## tax ptratio black lstat medv
## 1 1 1 1 1summary(boston_scaled)## crim zn indus
## Min. :-0.419367 Min. :-0.48724 Min. :-1.5563
## 1st Qu.:-0.410563 1st Qu.:-0.48724 1st Qu.:-0.8668
## Median :-0.390280 Median :-0.48724 Median :-0.2109
## Mean : 0.000000 Mean : 0.00000 Mean : 0.0000
## 3rd Qu.: 0.007389 3rd Qu.: 0.04872 3rd Qu.: 1.0150
## Max. : 9.924110 Max. : 3.80047 Max. : 2.4202
## chas nox rm age
## Min. :-0.2723 Min. :-1.4644 Min. :-3.8764 Min. :-2.3331
## 1st Qu.:-0.2723 1st Qu.:-0.9121 1st Qu.:-0.5681 1st Qu.:-0.8366
## Median :-0.2723 Median :-0.1441 Median :-0.1084 Median : 0.3171
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.:-0.2723 3rd Qu.: 0.5981 3rd Qu.: 0.4823 3rd Qu.: 0.9059
## Max. : 3.6648 Max. : 2.7296 Max. : 3.5515 Max. : 1.1164
## dis rad tax ptratio
## Min. :-1.2658 Min. :-0.9819 Min. :-1.3127 Min. :-2.7047
## 1st Qu.:-0.8049 1st Qu.:-0.6373 1st Qu.:-0.7668 1st Qu.:-0.4876
## Median :-0.2790 Median :-0.5225 Median :-0.4642 Median : 0.2746
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.6617 3rd Qu.: 1.6596 3rd Qu.: 1.5294 3rd Qu.: 0.8058
## Max. : 3.9566 Max. : 1.6596 Max. : 1.7964 Max. : 1.6372
## black lstat medv
## Min. :-3.9033 Min. :-1.5296 Min. :-1.9063
## 1st Qu.: 0.2049 1st Qu.:-0.7986 1st Qu.:-0.5989
## Median : 0.3808 Median :-0.1811 Median :-0.1449
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.4332 3rd Qu.: 0.6024 3rd Qu.: 0.2683
## Max. : 0.4406 Max. : 3.5453 Max. : 2.9865Then we use the quantile function to break the crime rates to 4 bins and categorize the crime rates to 4 different categories. Then we remove the original crime rate variuable and substitute it with the new categorial variables.
bins <- quantile(boston_scaled$crim)
crime <- cut(boston_scaled$crim, breaks = bins, include.lowest = TRUE, labels = c("low", "med_low", "med_high", "high"))
boston_scaled <- dplyr::select(boston_scaled, -crim)
boston_scaled <- data.frame(boston_scaled, crime)After that we divide the data to test and train sets, with 80 % of the data going to train set. We also save the correct classes from the test set and remove the crime variable from it.
n <- nrow(boston_scaled)
ind <- sample(n, size = n * 0.8)
train <- boston_scaled[ind,]
test <- boston_scaled[-ind,]Linear discriminant analysis
lda.fit <- lda(crime~., data=train)
classes <- as.numeric(train$crime)
plot(lda.fit, dimen=2, col=classes)
Predictions from the LDA
First we save the correct classess in the test set and remove it from the data. Then we predict the crime rate in the test set using the model from ther train set and cross-tabulate it with the correct classes.
correct_classes <- test$crime
test <- dplyr::select(test, -crime)
lda.pred <- predict(lda.fit, newdata=test)
table(correct = correct_classes, predicted = lda.pred$class)## predicted
## correct low med_low med_high high
## low 15 9 1 0
## med_low 4 12 2 0
## med_high 0 14 15 0
## high 0 0 1 29The model predicted quite well the crime categories. Probably with fewer classes, e.g. 3, the result would have been even better.
Euclidean distance and k-means clustering
data("Boston")
boston_scaled <- scale(Boston)
boston_euk_dist <- dist(boston_scaled)Then we can run the k-means clustering with 3 clusters and visualise the result with few relevant variables.
km <-kmeans(boston_scaled, centers = 3)
pairs(boston_scaled[,6:10], col = km$cluster)
It seems that 3 clusters might not be the best, so we can investigate the optimal number of clusters.
set.seed(123)
k_max <- 10
twcss <- sapply(1:k_max, function(k){kmeans(boston_scaled, k)$tot.withinss})
qplot(x = 1:k_max, y = twcss, geom = 'line')
It seems that 2 is the most optimal number of clusters, so let’s visualise that on the same variables.
km <-kmeans(boston_scaled, centers = 2)
pairs(boston_scaled[,6:10], col = km$cluster)
Looks better.